Questions and solutions on Probability

Questions and solutions on Probability

Question:

John looks at the clock at some arbitrary time between 4pm and 6 pm and notices that the angle between the two hands is an acute angle. He looks at the clock exactly 15 minutes later, what is the probability that during this observation also he sees that the angle between the hour and minute hands to be an acute-angle?

A.  1/2

B.  6/11

C.  13/24

D.  11/19

Correct Answer : (C)

Explanation:

At 4:00 pm the angle between the hands is 1200. The minutes hand is 1200 behind the hour hand. Now, let us first see when the angle will become an acute angle. The angle between he hands will be acute when the minutes hand gains more than 300. Or, after 30/330 * 60 minutes.

Somewhere between 4:05 and 4: 10.From now, the angle will be acute till the minute hand reaches a point where it is 90 degree ahead of the hour hand. Or, till the time the minute and gains 1800 (Going from 900 degree behind to 900 ahead of hour hand), the angle will be acute. Or, for a spell of 180/330 * 60 minutes, the angle will be acute. Or for a spell of 360/11 minutes, the angle will be acute. For a spell of 32 8/11 minutes the angle will be acute.

So, if John had seen the clock at a time from 4 pm to 5pm, he would have seen it during this spell.  Now, if his second viewing had also happened within this spell, he would have seen an acute angle second time.

Or, t, t + 15 should have both been within this 32 8/11 range for him to have seen two acute angle observations. Or, t should have happened at least 15 minutes before the angle become obtuse again. If we have the time range for it being acute as (x, x + 32 8/11), The question can be restated as if t belongs to this range, what it is the probability that t + 15 also belongs to this range.

For t + 15 to be within the range, t + 15 < x + 32 8/11, or t < 32 8/11 – 15 = 17 8/11

Probability should be 195/11 divided by 360/11 = 195/360 = 39/72 = 13/24

In the 5pm to 6pm range we get an identical probability. So the overall probability is 13/24

Answer Choice (C)

Level of Difficulty 2

CAT Challenging Question Mensuration

Here is an interesting question on mensuration

Question

Two horses are tied to opposite vertices of a rectangular field of length and breadth 6 and 6√2 meters. If they are tied with rope of length 6 meters, what is the common area that both horses can graze?
1. 6 π - 9√3
2. 12 π - 18√3
3. π - 2√3
4. 16 π -8√3
5. Cannot be calculated

People think that questions in 3-d (Cubes, Cuboid, cylinder, sphere etc) are more difficult in mensuration. In my view, the best questions are still from the 2-dimensional shapes. In 3-d, once you know the formula, you can pretty much solve everything. In 2-d, questions can include geometry and mensuration. Those are the best ones.

Anyways, happy solving. Will post solutions soon enough.


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CAT Practice Questions and Problems

Have provided below links to web pages in Ascent Education from where you can access sample problems and practice questions in Quant section, Verbal section and the DI and LR Section of the Computer Based Online CAT. The links are organized on a month wise basis.

August 2003 : You will find many Number Theory, Number Properties and Averages questions here.